Level 1 — RecognitionCalculus III — Sequences & Series

Calculus III — Sequences & Series

20 minutes30 marksprintable — key stays hidden on paper

Level 1 Paper — Recognition

Time limit: 20 minutes Total marks: 30


Section A — Multiple Choice (1 mark each)

Select the single best answer.

Q1. The sequence an=3n2+12n2+5a_n = \dfrac{3n^2+1}{2n^2+5} converges to: (a) 00 (b) 32\dfrac{3}{2} (c) \infty (d) 11

Q2. The geometric series n=0(23)n\displaystyle\sum_{n=0}^{\infty} \left(\tfrac{2}{3}\right)^n converges to: (a) 23\dfrac{2}{3} (b) 32\dfrac{3}{2} (c) 33 (d) diverges

Q3. The pp-series n=11np\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^p} converges if and only if: (a) p>1p>1 (b) p1p\geq 1 (c) p<1p<1 (d) p>0p>0

Q4. By the Divergence Test, n=1nn+1\displaystyle\sum_{n=1}^{\infty} \dfrac{n}{n+1}: (a) converges to 11 (b) diverges (c) converges to 00 (d) test is inconclusive

Q5. For the Ratio Test, the series an\sum a_n converges absolutely if limnan+1an=L\displaystyle\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right| = L with: (a) L>1L>1 (b) L=1L=1 (c) L<1L<1 (d) L=0L=0 only

Q6. The Maclaurin series of exe^x is: (a) n=0xnn!\displaystyle\sum_{n=0}^{\infty}\dfrac{x^n}{n!} (b) n=0(1)nxn\displaystyle\sum_{n=0}^{\infty}(-1)^n x^n (c) n=0(1)nx2n(2n)!\displaystyle\sum_{n=0}^{\infty}\dfrac{(-1)^n x^{2n}}{(2n)!} (d) n=1xnn\displaystyle\sum_{n=1}^{\infty}\dfrac{x^n}{n}

Q7. The radius of convergence of n=0xn2n\displaystyle\sum_{n=0}^{\infty}\dfrac{x^n}{2^n} is: (a) 11 (b) 22 (c) 12\tfrac12 (d) \infty

Q8. The Alternating Series (Leibniz) Test requires the terms bnb_n to be: (a) increasing and 0\to 0 (b) positive, decreasing, and 0\to 0 (c) bounded only (d) \to \infty

Q9. A series that converges but whose absolute-value series diverges is called: (a) absolutely convergent (b) divergent (c) conditionally convergent (d) geometric

Q10. The telescoping series n=1(1n1n+1)\displaystyle\sum_{n=1}^{\infty}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right) converges to: (a) 00 (b) 11 (c) 22 (d) diverges


Section B — Matching (1 mark each, Q11 total 5 marks)

Q11. Match each series in Column A to its correct classification in Column B.

Column A Column B
(i) 1n\sum \frac{1}{n} (P) converges (geometric, $
(ii) 1n2\sum \frac{1}{n^2} (Q) diverges (pp-series, p=1p=1)
(iii) (1)n+11n\sum (-1)^{n+1}\frac{1}{n} (R) converges (pp-series, p>1p>1)
(iv) (12)n\sum \left(\frac{1}{2}\right)^n (S) diverges (divergence test)
(v) n2n+1\sum \frac{n}{2n+1} (T) conditionally convergent

Section C — True/False WITH justification (2 marks each: 1 for T/F, 1 for reason)

Q12. Every bounded sequence converges. (T/F + justify)

Q13. If limnan=0\displaystyle\lim_{n\to\infty} a_n = 0, then an\sum a_n converges. (T/F + justify)

Q14. The Squeeze Theorem gives limnsinnn=0\displaystyle\lim_{n\to\infty}\dfrac{\sin n}{n}=0. (T/F + justify)

Q15. Absolute convergence implies convergence. (T/F + justify)

Q16. The Ratio Test with L=1L=1 proves the series diverges. (T/F + justify)

Q17. The Maclaurin series of cosx\cos x contains only even powers of xx. (T/F + justify)

Q18. A monotonic sequence that is bounded above must converge. (T/F + justify)


Answer keyMark scheme & solutions

Section A (1 mark each)

Q1 — (b) 3/23/2. Divide numerator and denominator by n2n^2: 3+1/n22+5/n232\frac{3+1/n^2}{2+5/n^2}\to \frac32. (1)

Q2 — (c) 33. Geometric with a=1a=1, r=23r=\tfrac23: sum =112/3=3=\frac{1}{1-2/3}=3. (1)

Q3 — (a) p>1p>1. Standard pp-series result via integral test. (1)

Q4 — (b) diverges. limnn+1=10\lim \frac{n}{n+1}=1\neq0; divergence test. (1)

Q5 — (c) L<1L<1. Ratio test convergence condition. (1)

Q6 — (a) xn/n!\sum x^n/n!. Definition of exe^x Maclaurin series. (1)

Q7 — (b) 22. x/2<1x<2|x/2|<1 \Rightarrow |x|<2. (1)

Q8 — (b) positive, decreasing, 0\to 0. Leibniz conditions. (1)

Q9 — (c) conditionally convergent. By definition. (1)

Q10 — (b) 11. Partial sum SN=11N+11S_N = 1-\frac{1}{N+1}\to 1. (1)

Section B

Q11 — (i)→(Q); (ii)→(R); (iii)→(T); (iv)→(P); (v)→(S). (1 each, 5 total) Reasoning: harmonic (p=1p=1) diverges; p=2>1p=2>1 converges; alternating harmonic converges but 1/n\sum 1/n diverges ⇒ conditional; r=12r=\tfrac12 geometric converges; n2n+1120\frac{n}{2n+1}\to\tfrac12\neq0 diverges.

Section C (2 marks each: 1 T/F, 1 reason)

Q12 — False. an=(1)na_n=(-1)^n is bounded but oscillates and does not converge. (1+1)

Q13 — False. Necessary not sufficient: 1/n\sum 1/n has an0a_n\to0 yet diverges. (1+1)

Q14 — True. 1nsinnn1n-\frac1n \le \frac{\sin n}{n}\le \frac1n, both bounds 0\to 0, so squeeze gives 00. (1+1)

Q15 — True. If an\sum|a_n| converges then an\sum a_n converges (standard theorem). (1+1)

Q16 — False. L=1L=1 is inconclusive; the test gives no information. (1+1)

Q17 — True. cosx=(1)nx2n(2n)!\cos x = \sum \frac{(-1)^n x^{2n}}{(2n)!}; all exponents even. (1+1)

Q18 — True. Monotone Convergence Theorem: bounded + monotone ⇒ convergent. (1+1)

[
  {"claim":"Q1 limit of (3n^2+1)/(2n^2+5) is 3/2","code":"n=symbols('n'); result = limit((3*n**2+1)/(2*n**2+5), n, oo) == Rational(3,2)"},
  {"claim":"Q2 geometric sum sum_{n=0}^inf (2/3)^n = 3","code":"n=symbols('n'); result = summation(Rational(2,3)**n,(n,0,oo)) == 3"},
  {"claim":"Q10 telescoping sum sum_{n=1}^inf (1/n-1/(n+1)) = 1","code":"n=symbols('n'); result = summation(1/n-1/(n+1),(n,1,oo)) == 1"},
  {"claim":"Q14 squeeze limit sin(n)/n -> 0","code":"n=symbols('n'); result = limit(sin(n)/n, n, oo) == 0"}
]